Vertical curve +0.5% and -3.5%, 1266 m at 780

Data: $g_1 = +0.5\% = +0.005$, $g_2 = -3.5\% = -0.035$ (a summit curve), IP at chainage 780.00 m, RL 1267.00 m. The curve passes through RL 1266.00 m at chainage 780 m (below the IP, as it must for a summit curve).

Length of the curve

$e = 1267.00 - 1266.00 = 1.00$ m, and for an equal tangent parabolic curve $e = \dfrac{(g_1 - g_2)L}{8}$:

$$L = \frac{8e}{g_1 - g_2} = \frac{8\times 1.00}{0.005 + 0.035} = 200.00\ \text{m}$$

Elements

• PVC chainage $= 780 - 100 = 680.00$ m; PVT chainage $= 880.00$ m
• RL of PVC $= 1267.00 - 0.005\times 100 = 1266.500$ m; RL of PVT $= 1267.00 - 0.035\times 100 = 1263.500$ m
• $r = \dfrac{g_2 - g_1}{L} = -0.000200$ per m

Levels at full stations (30 m pegs)

$$RL_x = RL_{PVC} + g_1x + \frac{r}{2}x^2$$

Highest point

$$x = \frac{-g_1}{r} = \frac{0.005}{0.000200} = 25.00\ \text{m}$$

Chainage $= 680.00 + 25.00 = 705.00$ m and RL $= 1266.562$ m.

Check: level at chainage 780 m (x = 100 m) $= 1266.000$ m = 1266.000 m as required.

Answer: L = 200 m; highest point at chainage 705.00 m, RL 1266.562 m.